$\newcommand\om\omega\newcommand\Om\Omega\newcommand\ep\epsilon$Let $(P_t)_{t\in T}$ be an exponential family over a separable complete metric space $(X,d)$, where $T$ is an open subset of $\mathbb R^k$ and $t=(t_1,\dots,t_k)$ is a [natural parameter][1]. So, for all $t\in T$, the density $p_t$ of $P_t$ with respect to Borel measure $\mu$ on $X$ is given by the formula 
$$p_t(x)=c(t)\exp\{t\cdot U(x)\}\tag1$$
for all $x\in X$, where 
$U\colon X\to\mathbb R^k$ is a Borel-measurable function, $\cdot$ denotes the dot product on $\mathbb R^k$, and 
$$c(t):=1\Big/\int_X\mu(dx)\exp\{t\cdot U(x)\}$$ 
is the normalizing factor. 
Assume that, for some (and hence for all) $x_0\in X$, the "first moment" 
$$m_t(x_0):=\int_X P_t(dx)d(x_0,x)=\int_X \mu(dx)p_t(x)d(x_0,x) $$
is finite for all $t\in T$. 

Then the parameterization map $U\ni t\mapsto P_t$ is continuous with respect to the Wasserstein-1 distance $W_1$. 

Indeed, by the [penultimate paragraph of the introduction section][2], if suffices to show that for each $t\in T$
$$\int_X \mu(dx)p_s(x)f(x)\to \int_X \mu(dx)p_t(x)f(x)$$
as $s\to t$, where $f\colon X\to\mathbb R$ such that $\sup_{x\in X}|f(x)|/(1+d(x_0,x))<\infty$ (for some (and hence for all) $x_0\in X$). 
This follows immediately by the [uniform integrability][3]. 

In turn, in view of (1), the uniform integrability here follows by the following argument: Take any $t=(t_1,\dots,t_k)$ in the open set $T$. Take any real $\ep>0$ such that $t+2\ep\om\in T$ for all $\om=(\om_1,\dots,\om_k)\in\Om:=\{-1,1\}^k$. Then for any $s=(s_1,\dots,s_k)\in T$ such that $\max_{j=1}^k|s_j- t_j|\le\ep$ we have 
\begin{align}\exp\{s\cdot u\}/\exp\{t\cdot u\}&=\exp\{(s-t)\cdot u\} \\ 
&\le\max_{\om\in\Om}\exp\{\ep\om\cdot u\} \\ 
&=\exp\Big\{\ep\sum_{j=1}^k|u_j|\Big\} \\ 
&<<\exp\Big\{2\ep\sum_{j=1}^k|u_j|\Big\} \\ 
&=\max_{\om\in\Om}\exp\Big\{2\ep\om\cdot u\Big\} \\ 
&\le\sum_{\om\in\Om}\exp\Big\{2\ep\om\cdot u\Big\};
\end{align}
the first inequality in the above display follows because the function $\exp$ is convex and the condition $\max_{j=1}^k|s_j- t_j|<\ep$ implies that the point $s-t$ is in the convex hull of the set $\{\ep\om\colon\om\in\Om\}$; here $a<<b$ mean that $a=o(b)$ as $\sum_{j=1}^k|u_j|\to\infty$. So, uniformly over all $s\in T$ such that $\max_{j=1}^k|s_j- t_j|\le\ep$ we have 
\begin{align}\exp\{s\cdot u\}<<\sum_{\om\in\Om}\exp\{( t+2\ep\om)\cdot u\},
\end{align}
and $t+2\ep\om\in T$ for all $\ep=(\ep_1,\dots,\ep_k)\in\Om:=\{-1,1\}^k$. Thus, the uniform integrability follows. $\Box$



  [1]: https://en.wikipedia.org/wiki/Exponential_family#Vector_parameter
  [2]: https://www.lpsm.paris/pageperso/bolley/wasserstein.pdf
  [3]: https://en.wikipedia.org/wiki/Uniform_integrability